12 UNBOUNDED NORMAL OPERATORS 425 (g) Suppose that Im(T) is closed in F. If M => Ker(T) is a closed subspace of E, show that T(M n dom(T)) is closed in F. (Consider the restriction of T to M n dom(7) as an unbounded operator from M to F, and use (f).) (h) Let N be a closed subspace of F such that N n Im(7) = {0}. If N -f lm(T) is closed in F, then Im(r) is closed in F, (Consider the operator T2 from E x N to F defined on dom(7") x N by T2 • (x, y) = T - x -f- y, and remark that y(T) ;> y(T2).) In particular, if lm(T) has finite codimension in F, then Im(T) is closed in F. 2. Let E, F, H be three Hilbert spaces, Tan unbounded operator from E to F, and U an unbounded operator from H to E. Suppose that T is closed, dom(r) is dense in E, Ker(7") is finite-dimensional and Im(:T) is closed in F. (a) Show that if U is closed then so is TU (defined on dom(l/) n t/""1 (dom(T)).) (Suppose that zn e H tends to z and that TU - zn tends to y e F. Using the relation y(T) > 0 (Problem 1), show that there exists a sequence (*„) in Ker(T) such that U • zn + xn has a limit x in E. Use the fact that Ker(T) is locally compact to show that the sequence (xn) must be bounded (argue by contradiction) and then pick a conver- gent subsequence of (#„).) (b) If U is closed and Im(C/) is closed in E, then Im(TU) is closed in F. (Use Problem l(g) and (5.9.2).) (c) Suppose that Ker(t/) is finite-dimensional. Show that Ker(7T7) is finite-dimen- sional and that dim(Ker(7I/)) = dim(Ker(t/)) + dim(Im(£7) n Ker(D). (d) Suppose that U is closed, dom(U) dense in H and lm(U) of finite codimension in E (which implies, by virtue of Problem l(h), that lm(U) is closed in E). Show that dom(TU) is dense in H. (Consider the orthogonal supplement Hx of Ker(C7) in H, and the restriction C/i of U to the dense subspace dom(£7) n HI of Hj ; observe that U"1 is continuous on Im(C/i) == Im(C7) and that dom(r) n Im(U) is dense in Im(£/).) (e) Suppose that Im(r) and Im(t/) have finite codimension in F and E, respectively, and let v be the codimension of Im(£/) n Ker(r) in Ker(7"). Show that lm(TU) has finite codimension in F and that codim(Im(7T/)) = codim(Im(r)) 4- codim(Im(£/)) - v. (Remark that E is the direct sum of Im(t/), a supplement NI of Im((7) n Ker(J') in Ker(r), and a subspace N2 of finite dimension contained in dom(!T), and that the restriction of T to N2 is injective.) 3. Let E, F be two Hilbert spaces. An unbounded operator T from E to F is said to be an operator with index (Section 15.11, Problem 22) if T is closed, dom(T') dense in E, Ker(T") finite-dimensional and Im(!T) of finite codimension in F (in which case T(E) = lm(T) is closed in F by Problem l(h)). The index ofTis defined as the number i(T) - dim(Ker(D) - codim(Im(D). (a) If Tis an index operator from E to F, then T* is an index operator from F to E, andi(r*) = -/(D (use Problem l(e)). (b) Deduce from Problem 2 that if U : H -*• E and T : E -> F are index operators, then TU : H -> F is an index operator and i(TU) = i(T) + i(U). (c) Let TI be an unbounded operator from E to F, which extends Tand is such that dom(7\) = dom( J") © M, where M is a finite-dimensional subspace of E. If T isn E0, to — ToSj;1 on EI, and to —Sn,iT0Snl onEnfor